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What is the Power Factor of CR?
Fig shows the circuit. Alternating voltage applied to the circuit,
V = V0sinωt
At any moment if Q be the charge stored in the capacitor C, then the effective voltage in the circuit,
From equations (1) and (3) we conclude the followings.
i) The current I leads the voltage V by a phase angle θ,
where tanθ = \(\frac{\sin \theta}{\cos \theta}\) = \(\frac{1 / \omega C}{Z} \times \frac{Z}{R}\) = \(\frac{1}{\omega C R}\),
i.e., θ = tan-1\(\left(\frac{1}{\omega C R}\right)\)
This phase relation is shown in Fig. We know that in a pure resistive circuit, V and I are in the same phase, i.e., phase difference, θ = 0. On the other hand, in a pure capacitive circuit, I always leads V by θ = 90°. So, in a CR circuit, 1 should lead V by 0 < θ < 90°.
ii) Z plays the same role as R in a pure resistive circuit. This Z is know as the impedance of the circuit.
Impedance, Z = \(\sqrt{R^2+\left(\frac{1}{\omega C}\right)^2}\) = \(\sqrt{R^2+X_C^2}\) …… (4)
where XC = \(\frac{1}{\omega C}\) = capacitive reactance.
Impedance in a CR circuit is the effective resistance of the circuit arising from the combined
effects of ohmic resistance and capacitive reactance.
Fig. shows the impedance triangle for the circuit.
Power in the circuit : Here, power factor of the circuit, cosθ = \(\frac{R}{Z}\)
So, P = \(\frac{1}{2} V_0 I_0 \cos \theta\) = \(\frac{1}{2}\left(I_0 Z\right) I_0 \cdot \frac{R}{Z}\) = \(\frac{1}{2} I_0^2 R\) = \(\left(\frac{I_0}{\sqrt{2}}\right)^2 R\)
∴ P = \(I_{\mathrm{rms}}^2 \cdot R\)
Hence, power is dissipated only in the resistance R. Current through the capacitor C is wattless.